25 × 4 = 100 - the number of all candies
2/5 of 100 is
![\dfrac{2}{5}\cdot100=\dfrac{200}{5}=40](https://tex.z-dn.net/?f=%5Cdfrac%7B2%7D%7B5%7D%5Ccdot100%3D%5Cdfrac%7B200%7D%7B5%7D%3D40)
The third model is shaded to show the total number of caramel candies in 4 bags.
Answer:14 dimes 10 nickels
Step-by-step explanation:
24 coins total
14 X.10 = 1.40
10 x .05 = .50
1.40 + .50 = 1.90
Answer:
The inequality for Part A: −6 °F < −2 °F
Step by step-
The reason for that us because the -2 is closer to zero on the number line
To complete the square you halve the coefficient of the x term and square it. Half of 14 is 7 and 7² is 49. So we add 49 and subtract 49, which means we are not changing the value of the quadratic. So we have x²+14x+49-49+2. This can be written: (x²+14x+49)-49+2, which is (x+7)²-47, which is answer a.
Answer:
The percentage of students who scored below 620 is 93.32%.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
![\mu = 500, \sigma = 80](https://tex.z-dn.net/?f=%5Cmu%20%3D%20500%2C%20%5Csigma%20%3D%2080)
Percentage of students who scored below 620:
This is the pvalue of Z when X = 620. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{620 - 500}{80}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B620%20-%20500%7D%7B80%7D)
![Z = 1.5](https://tex.z-dn.net/?f=Z%20%3D%201.5)
has a pvalue of 0.9332
The percentage of students who scored below 620 is 93.32%.